Optimizing with Derivatives I think

[michaeldw]

Homework Statement

You're on the moon and you need a bit from the base 900 feet away so you can continue your experiment. The base has a roover/launcher used for retrieving the bit. The roover can move a max dist of 600 feet. The launcher is controlled by angle. At this time the angle is inclined at 70 degrees. It cost $375 per foot to move the roover, and $20,000 per degree to change the launch angle. What are the optimal settings (what angle should I launch from where) to minimize cost?

acceleration due to grav on moon = 5.31 ft/s^2
initial velocity of bit leaving launcher = 75 ft/s

Homework Equations

R = V^2(sin(2x))/g ; where R = total distance, V= Velocity, x= theta (some angle) and g= gravitational acceleration.

The Attempt at a Solution

I made an equation for constraint and pne for the optimaization.

Constraint:
900 = [[(75^2)*sin(2x)]/5.31]+ y ; where y = distance roover moves and x = theta

optimization
minCost = [20000*|70-x|]+2y(375) ; |...| = absolute value, 2y because the roover has to return to base.

I remember doing problems like this in HS but I need help with this one. I solved for y in the constraint equation, then substiituted it in for y in the optimization equation. Here's what I got:

minCost= [-7.945*(10^5)*sin(2x)+675000+20000*|x-70|]

Now if I am correct I get the 1st derivative of this and I will get the extrema or something like that which will be the optimal angle. Then I plug that into the first equation and get y...

But how do I find the first derivative. This might not even be the right method so please help thatnks.

 

[michaeldw]

whoops wrong section

 

[Architeuthis Dux]

Your approach to solving this problem using derivatives is correct. The first step is to set up the equations for the constraint and optimization. From there, you can use the constraint equation to solve for one variable in terms of the other. In this case, you solved for y in terms of x.

To find the minimum cost, you need to take the derivative of the optimization equation with respect to x, set it equal to 0, and solve for x. This will give you the value of x that minimizes the cost. Once you have that value, you can plug it back into the constraint equation to solve for y.

To find the derivative of the optimization equation, you can use the chain rule and product rule. Since there is an absolute value in the equation, you will need to use the derivative of an absolute value function as well. Once you have the derivative, set it equal to 0 and solve for x.

Remember to check the second derivative to make sure the value of x you found is a minimum and not a maximum. If the second derivative is positive, then the value of x is a minimum and you can proceed with solving for y. If the second derivative is negative, then the value of x is a maximum and you will need to re-evaluate your equations to find the minimum cost.

I hope this helps and good luck with your optimization problem!